Ultracaps tested for excess energy

[MileHigh]

Gadget:

The Bcap Will react Differently with Frequency charge  just like an led will react differently but you get the same effect with less power from a source

I think that it is fair to say that this is what you hope.  You think that pulsing DC into the ultracapacitor will give you more energy in the capacitor than you put into it.  In other words the process of pulsing DC current into the ultracap is an over unity process as compared to charging it with pure DC current which is not.

same with Battery caps only they don't require the amount of joules the constant current supply gives them

The mathematics and the reality say this is not the case.  There is no relationship between an ultracap and an LED.  Whether you put 100 Joules of energy into the cap from a constant DC current, or you do it with pulsing DC, you will end up putting the same 100 Joules of energy into the capacitor.

Here is just an approximate breakdown:

1.  100 Joules of energy made available to put into the ultracapacitor from DC or pulsing source.
2.  At the end of charging the capacitor you have 99.8 Joules of energy in the cap, you lost some energy from the equivalent series resistance of the cap.  0.2 Joules became heat during the charging process.
3.  When you discharge the ultracap you get 99.6 Joules of energy out of the capacitor and 0.2 Joules of energy became heat during the discharging process due to the equivalent series resistance.

MileHigh

[gadgetmall]

Gadget:

I think that it is fair to say that this is what you hope.  You think that pulsing DC into the ultracapacitor will give you more energy in the capacitor than you put into it.  In other words the process of pulsing DC current into the ultracap is an over unity process as compared to charging it with pure DC current which is not.

The mathematics and the reality say this is not the case.  There is no relationship between an ultracap and an LED.  Whether you put 100 Joules of energy into the cap from a constant DC current, or you do it with pulsing DC, you will end up putting the same 100 Joules of energy into the capacitor.

Here is just an approximate breakdown:

1.  100 Joules of energy made available to put into the ultracapacitor from DC or pulsing source.
2.  At the end of charging the capacitor you have 99.8 Joules of energy in the cap, you lost some energy from the equivalent series resistance of the cap.  0.2 Joules became heat during the charging process.
3.  When you discharge the ultracap you get 99.6 Joules of energy out of the capacitor and 0.2 Joules of energy became heat during the discharging process due to the equivalent series resistance.

MileHigh

Calculate how many Joules are in an aa battery @ 1.450 volts @2500mah  or 2.5 amps  and then please calculate  joules in a charged bcap 2.7 volts and you know the specs 3500amps  please ?
And what i said was that the Bcap CHARGED OVER THE UNITY OF THE BATTERY as stated in wiki . that is a fact battery 1.4 volts bcap charged to 2.6volts . energy or not, volts are still 1.4 volts in battery and bcap charged from .5 volts to 2.6 volts .. bcap can light up nichrome wire , aa battey can not .. cap can run more than the aa battery can for a longer period than aa battery after charged .Bcap can run a 5 watt red led on its own after recharge all day and night aa battery on its own wont light it up at all . Normal ? No.

[gadgetmall]

Right . you wont answer because this will prove i am right .
i know the battery is 3.5 watts
and the cap is 9100 watts

but i dont know how to calculate joules from this

[MileHigh]

Gadget:

The energy in the cap is 1/2 C v-squared

E = 1/2 * 350 * 2.7-squared

E = 1275.75 Joules


Now let's forget about the ultracapacitor for a second and look at power levels:

A watt is one Joule per second.

Now, let's look at how much power is being expended if you imagine 3500 amps at 2.7 volts.  3500 x 2.7 = 9450 watts = 9450 Joules per second.

The capacitor voltage goes DOWN as it discharges.  That means that the current goes DOWN as it discharges through a resistive load, and the power level goes DOWN as it discharges through a resistive load.

The only way to have the capacitor power dissipation stay CONSTANT is to discharge it through a VARIABLE resistor.

Let's imagine that you have this variable resistor to keep the power constant.

If you want the capacitor to discharge at a CONSTANT power level of 9450 watts = 9450 Joules per second, and you have 1275.75 Joules of energy available in the capacitor,  then it will take 1275.75/9450 = 0.135 seconds.

Poynt did the battery energy calculation and if I recall the battery has enough energy to charge the 350 farad ultracap to 2.7 volts about 8 times.

Therefore if you imagine that the battery fills the ultracap about 8 times, and you can discharge the cap at a constant power level of 9450 watts, then you can get about (8 x 0.135) = 1.08 seconds of total discharge time from the ultracap at this very high power level.

In other words, all of the battery energy could be consumed in about one second at a power level of 9450 watts using the ultracap to do this in eight separate discharges.

Calculate how many Joules are in an aa battery @ 1.450 volts @2500mah  or 2.5 amps  and then please calculate  joules in a charged bcap 2.7 volts and you know the specs 3500amps  please ?

I suggest that you work with the people in the JT thread for your project to understand energy and power.  In your question above the 3500 amps is meaningless and is just the short circuit spec for the ultracap in the datasheet.

It is very obvious that you DO NOT UNDERSTAND how energy, time, and power work together.  If you want to be serious about your project then in my opinion you have no choice, you HAVE TO UNDERSTAND these concepts.  It's what the whole web site is about.

Your example with the nichrome wire is totally meaningless.  If you work with the people on your thread to understand these concepts then you would also realize that your example is totally meaningless.

I challenge you to work with the people on your thread so that at least all of you understand what you are doing and what you are trying to accomplish.  I am not going to be your teacher, the people working with you on your project can be your teachers and you can try to teach yourself also.

You are not alone here.  On the Rosemary Ainsley thread the main component in her circuit was an inductor, and Rosemary had no understanding of how an inductor worked when I started speaking to her.  Life is a continuous learning process and you have no choice but to learn if you want to be an active member of the project.  This probably applies to other members of the JT thread that want to build the PCB and make measurements.

MileHigh

[electricme]

@Gadget,
Ha ha, welcome to the cartoonists corner, nice feeling isn't it.
I like the colour cartoon, nice one Al.


@ Paul,
Well done, congratulations on the successful running of the data logger against Gadget's Cap test.

When the datalogger circuit and the software are ready, I would love a copy please.

Thankyou.

jim